I was trying to prove that as $k\to1^-$,
$$ \int_0^1{\mathrm dx\over\sqrt{1-x^2}\sqrt{1-k^2x^2}}\sim\frac12\log{1\over1-k} $$
Since the original integral is difficult to work with, I split it:
$$ \int_0^1{\mathrm dx\over\sqrt{1-x^2}\sqrt{1-k^2x^2}}=\int_0^k+\int_k^1 $$
in which it is evident that
$$ \begin{aligned} \int_0^k{\mathrm dx\over\sqrt{1-x^2}\sqrt{1-k^2x^2}} &=\int_0^k{\mathrm dx\over1-x^2}\cdot{\sqrt{1-x^2}\over\sqrt{1-k^2x^2}} \\ &=\frac12\int_0^k{\mathrm dx\over1-x}\cdot{\sqrt{1-x^2}\over\sqrt{1-k^2x^2}}+\mathcal O(1) \\ &=\frac12\int_0^k{\mathrm dx\over1-x}+\mathcal O\left(\int_0^k{1-k\over1-x}\mathrm dx\right)+\mathcal O(1) \\ &=\frac12\log{1\over1-k}+\mathcal O(1) \end{aligned} $$
I wonder how to show that $\int_k^1$ converges to zero as $k\to1^-$ so that I can complete the proof.
P.S. Alternative methods are welcome since I have been stuck on this for so much time :D
If $0<k<x<1$, then \begin{align*} \sqrt {1 - x^2 } \sqrt {1 - k^2 x^2 } = \sqrt {1 - x} \sqrt {1 + x} \sqrt {1 - kx} \sqrt {1 + kx} \ge \sqrt {1 - x} \sqrt {1 - k} . \end{align*} Hence, $$ 0 \leq \int_k^1 {\frac{{dx}}{{\sqrt {1 - x^2 } \sqrt {1 - k^2 x^2 } }}} \le \frac{1}{{\sqrt {1 - k} }}\int_k^1 {\frac{{dx}}{{\sqrt {1 - x} }}} = 2. $$ This bound together with your estimate for the integral between $0$ and $k$ imply the desired asymptotics.